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Prove that ${\tan ^{ – 1}}\frac{1}{4} + {\tan ^{ – 1}}\frac{2}{9} = {\sin ^{ – 1}}\frac{1}{{\sqrt 5 }}$. [NCERT,Exemplar.2.3,Q.16,Page.36]
We have, ${\tan ^{ – 1}}\frac{1}{4} + {\tan ^{ – 1}}\frac{2}{9} = {\sin ^{ – 1}}\frac{1}{{\sqrt 5 }}$ ….(i) Let ${\tan ^{ – 1}}\frac{1}{4} = x$ $ \Rightarrow $$\tan x = \frac{1}{4}$
$ \Rightarrow $${\tan ^2}x = \frac{1}{{16}}$ $ \Rightarrow $${\sec ^2}x – 1 = \frac{1}{{16}}$
$ \Rightarrow $${\sec ^2}x = 1 + \frac{1}{{16}} = \frac{{17}}{{16}}$
$ \Rightarrow $$\frac{1}{{{{\cos }^2}x}} = \frac{{17}}{{16}}$
$ \Rightarrow $${\cos ^2}x = \frac{{16}}{{17}}$ $ \Rightarrow $$\cos x = \frac{4}{{\sqrt {17} }}$
$ \Rightarrow $${\sin ^2}x = 1 – {\cos ^2}x = 1 – \frac{{16}}{{17}} = \frac{1}{{17}}$
$ \Rightarrow $$\sin x = \frac{1}{{\sqrt {17} }}$
…..(ii) Again, let ${\tan ^{ – 1}}\frac{2}{9} = y$ $ \Rightarrow $$\tan y = \frac{2}{9} \Rightarrow {\tan ^2}y = \frac{4}{{81}}$
$ \Rightarrow $${\sec ^2}y – 1 = \frac{4}{{81}}$ $ \Rightarrow $${\sec ^2}y = \frac{4}{{81}} + 1 = \frac{{85}}{{81}}$
$ \Rightarrow $${\cos ^2}y = \frac{{81}}{{85}} \Rightarrow \cos y = \frac{9}{{\sqrt {85} }}$
$ \Rightarrow $${\sin ^2}y = 1 – {\cos ^2}y = 1 – \frac{{81}}{{85}} = \frac{4}{{85}}$
$ \Rightarrow $$\sin y = \frac{2}{{\sqrt {85} }}$
……(iii) We know that, $\sin (x + y) = \sin x \cdot \cos y + \cos x \cdot \sin y$
$ = \frac{1}{{\sqrt {17} }} \cdot \frac{9}{{\sqrt {85} }} + \frac{4}{{\sqrt {17} }} \cdot \frac{2}{{\sqrt {85} }}$
$ = \frac{{17}}{{\sqrt {17} \cdot \sqrt {85} }} = \frac{{\sqrt {17} }}{{\sqrt {17} \cdot \sqrt 5 }} = \frac{1}{{\sqrt 5 }}$
$ \Rightarrow $$(x + y) = {\sin ^{ – 1}}\frac{1}{{\sqrt 5 }}$
$ \Rightarrow $${\tan ^{ – 1}}\frac{1}{4} + {\tan ^{ – 1}}\frac{2}{9} = {\sin ^{ – 1}}\frac{1}{{\sqrt 5 }}$ Hence proved.
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